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Proof step 2 from proof subset-antisym (previous step)

bag1 ⊆ bag0, bag0 ⊆ bag1 ⊦ bag0 = bag1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ bag0 = bag1 ↔ (∀ a. count(a, bag0) = count(a, bag1))
Proof step 4:
bag1 ⊆ bag0, bag0 ⊆ bag1 ⊦ ∀ a. count(a, bag0) = count(a, bag1)

Rule argument

spec: bag-basic inst: name: extension seq: ⊦ bag0 = bag1 ↔ (∀ a. count(a, bag0) = count(a, bag1)) subst: [bag1, bag0] → [bag1, bag0] flags: rewrite

Additional information

Used terms: bag1, bag0